Magnetism and Matter: Class 12 Physics Chapter 5 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

110 min
2 simulations
17 practice questions
33 MCQs

Is chapter mein aap kya seekhenge

Bar magnet ke field ko field lines se describe karo, aur unki chaar properties se tay karo ki koi field-line picture possible hai ya nahi
Ampère ke circulating currents se samjhao ki bar magnet solenoid jaisa behave kyun karta hai, aur finite solenoid ka door ka axial field derive karo
Uniform field mein magnetic dipole ka torque aur potential energy calculate karo, aur stable aur unstable equilibrium pehchaano
Electric-dipole analogy se short magnet ke axial aur equatorial fields nikaalo
Gauss's law for magnetism state karo aur use magnetic monopoles ke na hone se jodo
Magnetisation, magnetic intensity, susceptibility aur permeability define karo, aur kisi material ke andar H, M aur B calculate karo
Materials ko diamagnetic, paramagnetic ya ferromagnetic mein classify karo aur har ek ko atomic level par samjhao, domains aur Curie temperature ke saath

NCERT Physics Class 12, Chapter 5, pp. 136–153 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

5.1 Introduction

Apne phone par map app kholo aur ek hi jagah par dheere-dheere ghoomo: screen par chhota arrow tumhare saath ghoomta hai. Kuch millimetre ka ek sensor chip Earth ka magnetic field measure kar raha hai. Magnetism aise gadgets se kahin aage tak jaata hai. Electrons aur atomic nuclei chhote-chhote magnets ki tarah behave karte hain, Earth aur Sun ke apne magnetic fields hain, aur galaxy ke stars ke beech ki patli gas mein bhi hazaaron light years tak phaile weak fields hain.

Log magnetism se uski explanation milne se bahut pehle mil chuke the. Greek writers ne ek kaale iron ore ke baare mein likha, jise ab magnetite ya lodestone kehte hain, jo iron ko kheenchta hai; Magnesia district, jahan yeh khoda jaata tha, se hi humein magnet shabd mila. Chapter 4 mein tumne jo bhi magnetic field dekha, woh sab moving charges se aaya tha. Yahan hum khud magnets ki baat karenge, aur yeh dekhenge ki magnetic field lagane par har tarah ka matter kaise respond karta hai (चुंबकत्व एवं द्रव्य).

Pichhli classes mein tumne magnets ke kuch rules seekhe the. Yahan woh chaar headings mein grouped hain; baaki chapter samjhata hai ki har rule kahan se aata hai.

Direction. Ek bar magnet ko dhaage se latkao, to woh tab tak ghoomta hai jab tak lagbhag north–south mein na aa jaaye. Jo end geographic north ki taraf rukta hai, use magnet ka north pole (N) kehte hain aur doosre end ko south pole (S). Magnet aisa isliye karta hai kyunki Earth khud magnetised hai: zameen ke paas uska field lagbhag geographic south se geographic north ki taraf chalta hai.
Forces. Do N poles ek doosre ko repel karte hain, do S poles repel karte hain, aur ek N pole aur ek S pole attract karte hain: like poles repel, unlike poles attract.
Pairs. Aaj tak kisi ne akela N ya S pole, yaani magnetic monopole, nahi banaya. Magnet ko kaise bhi todo, har tukde mein dono poles hote hain.
Materials. Permanent magnets iron aur uske jaise kuch metals, cobalt aur nickel, ya unke alloys aur compounds se bante hain; zyadatar doosre substances magnet ko bas bahut halka sa respond karte hain.

3/4Har tukda phir se apne N aur S ke saath poora magnet hai.

Fig. 5.1Magnet ko kaatne se uske poles kabhi alag nahi hote: har tukda, ek atom tak, N aur S wala poora magnet hota hai.

Ek horizontal bar magnet jiska S end blue aur N end orange hai, dono ends par iron filings ki chhoti lakeerein hain. Step 1: dono poles ke saath poora magnet. Step 2: ise beech se kaata gaya hai; dono halves alag hote hain aur har ek ka apna S half aur N half dikhta hai, aur chaaron ends par filings jama hoti hain. Step 3: har half ko phir kaata jaata hai, jisse chaar chhote magnets bante hain, har ek par pehle S phir N likha hai. Step 4: aakhri tukde ke neeche ek magnifier ek atom ko electron orbit ki tarah dikhata hai, jiske magnetic moment ka arrow magnet ke saath point karta hai, uske left mein S aur right mein N hai.

Pehle do rules ko neeche ke simulation mein test karo. Magnet A dhaage se latka hai, isliye woh free hokar ghoom sakta hai; magnet B ko drag, turn aur flip kiya ja sakta hai. Grey arrow geographic north dikhata hai.

SimulationLatka hua magnet aur haath mein pakda magnet

Ek stage jismein Earth ka field on hai aur kone mein North arrow hai. Magnet A left mein dhaage par latka hai aur settle hone tak ghoomta hai; magnet B, ek weaker bar magnet, right mein seedha khada hai, uska N end neeche hai, aur use drag, turn ya flip kiya ja sakta hai. Arrows dikhate hain ki dono magnets attract kar rahe hain ya repel, aur field lines on ki ja sakti hain.

Fig. 5.2Magnet A apne dhaage par free ghoomta hai; magnet B ko drag, turn ya flip karo aur dekho A kaise respond karta hai.

Try karo

Play dabao, B ko wahin rehne do aur dekho A kaise settle hota hai; ismein kuch seconds lagte hain. North arrow ke comparison mein A ka N end kis taraf point karta hai?
B ko A ki taraf drag karo jab tak B ka N end A ke N end ke bilkul paas na aa jaaye. A kis taraf ghoomta hai, aur A ka kaunsa end aakhir mein B ki taraf hota hai?
Flip B dabao aur B ko wapas usi jagah lao. Ab kya alag hai?

Timeline

Lodestone se superconductors tak

c. 600 BCE
Lodestone ka zikr
Greek writers likhte hain ki lodestone iron ko attract karta hai.
1088
Needle compass
Chinese scholar Shen Kuo ek magnetised needle describe karte hain jo south–north point karti hai.
1600
Earth ek magnet hai
William Gilbert ki De Magnete kehti hai ki poori Earth ek bahut bada magnet hai.
1820
Currents fields banate hain
Oersted dekhte hain ki current compass ko ghuma deta hai; Ampère propose karte hain ki saara magnetism circulating currents se aata hai.
1845
Diamagnetism
Faraday paate hain ki bismuth aur kai doosre substances strong magnet se halke se repel hote hain.
1895
Curie ke measurements
Pierre Curie dikhate hain ki temperature ke saath paramagnetism kaise kamzor hota hai aur garam hone par iron apna magnetism kaise kho deta hai.
1907
Domains
Pierre Weiss line-up hue atomic magnets ke regions se ferromagnetism samjhate hain.
1933
Meissner effect
Meissner aur Ochsenfeld paate hain ki superconductor magnetic field ko apne andar se bahar dhakel deta hai.

Pehla rule thoda dhyaan se dekhne laayak hai (Fig. 5.3). Agar Earth ka bahar wala field south se north ki taraf point karta hai, to geographic north ke paas wala magnetic pole compass ke N end ko attract karta hai. Isliye magnetically woh south pole hi hona chahiye.

3/3Field axis se tirche bar magnet ke field jaisa dikhta hai, jiska S end north ki taraf hai. Compass needles field lines ke saath line up hoti hain.

Fig. 5.3Earth ke bahar field southern hemisphere se northern hemisphere ki taraf chalta hai, maano andar ek tircha bar magnet baitha ho jiska S end Arctic ke neeche hai.

Ek circle Earth ko dikhata hai, jismein dashed vertical rotation axis hai jiske upar geographic N aur neeche geographic S likha hai, aur ek dashed equator hai. Step 1 mein sirf Earth aur uska axis hai. Step 2 mein field lines judti hain jo Earth ke southern half se nikalti hain, space mein loop banakar northern half mein wapas ghusti hain, aur loops ke bahar wale hisson par arrows north ki taraf point karte hain. Step 3 mein rotation axis se lagbhag das degree tircha ek dotted magnetic axis judta hai, uske saath ek imaginary bar magnet jiska S end geographic north aur N end geographic south ki taraf hai, aur surface par teen compass needles field ke saath line up hoti hain: magnetic equator par level, northern hemisphere mein N end zameen ki taraf neeche jhukta hai aur southern hemisphere mein zameen se upar uthta hai.

Warm-up

3 mein se 1

Freely latka bar magnet apna N end geographic north ki taraf karke rukta hai. Arctic ke neeche kis tarah ka magnetic pole hai?

Koi pole nahi, sirf rotation axisNorth magnetic pole, kyunki woh north mein haiSouth magnetic pole, kyunki woh N end ko attract karta hai

Answer dekho

Answer (C) South magnetic pole, kyunki woh N end ko attract karta hai

Unlike poles attract karte hain. Magnet ka N end Arctic ki taraf kheencha jaata hai, isliye wahan ka pole south pole ki tarah behave karta hai. Use 'north magnetic pole' kehna map par uski jagah batata hai, uska magnetic type nahi.

5.2 Bar magnet

Iron filings chhidke card ko thapthapao jab uske neeche ek chhota bar magnet rakha ho, aur ek pattern ubhar aata hai. Har filing field se magnetised hoti hai aur ghoomkar uske saath line up ho jaati hai, aur paas ki filings end to end judkar chains bana leti hain. Chains magnet ke dono ends par sabse ghani hoti hain aur ek end se doosre end tak arch banati hain. Ise Chapter 1 ke electric dipole ke field wali drawing ke saath rakho, to samanta saaf dikh jaati hai. Magnet ke ends do opposite charges ki tarah kaam karte hain, aur poora magnet ek magnetic dipole (चुंबकीय द्विध्रुव) hai.

Try karke dekho

Compass se field lines plot karo

Tumhe chahiye

ek bar magnet
ek chhota plotting compass
A4 paper ki ek sheet aur ek pencil
sticky tape

Pehle guess karo

Tumhe kya lagta hai, tumhari plot ki hui lines kahan sabse paas-paas hongi?

magnet ke beech ke paasmagnet ke dono ends ke paashar jagah barabar doori par

Paper ko table par tape karo, iron ki cheezon se door. Magnet ko beech mein rakho, uski outline banao aur N aur S mark karo.
Compass ko N end ke ek kone ke bilkul paas rakho. Needle ke N end ki tip par ek dot lagao.
Compass ko aise khiskao ki needle ki tail us dot par aa jaaye, aur tip par naya dot lagao. Yeh tab tak dohrao jab tak tum magnet ya paper ke edge tak na pahunch jao.
Dots ko ek smooth curve se jodo aur ek arrow banao jo dikhaye ki needle kis taraf point kar rahi thi.
Magnet ke aas-paas doosre points se shuru karo jab tak aath se das lines na ban jaayein.

Abhi nahi kar sakte? Simulation try karo · Fig. 5.5

5.2.1 Magnetic field lines

Magnetic field line (चुंबकीय क्षेत्र रेखा) ek aisa curve hai jo har point par wahan ke $𝐁$ ki direction mein chalta hai. Aisi lines ka set ek invisible vector field ko ek picture mein badal deta hai jise tum ek nazar mein padh sakte ho (Fig. 5.4). Inke chaar rules hain.

Yeh closed loops hoti hain. Magnet ke bahar lines N se S ki taraf chalti hain; magnet ke andar woh S se wapas N tak chalti rehti hain. Koi line kahin shuru ya khatam nahi hoti. (Chapter 1 ke electric dipole se compare karo, jiski lines ke ends hote hain: charges par.)
Tangent direction batata hai. Kisi bhi point par $𝐁$ us point se guzarne wali line ke tangent ke saath hota hai, aur arrow ki direction mein point karta hai. Wahan rakha chhota compass is tangent ke saath line up hota hai.
Crowding strength batati hai. Jahan lines paas-paas hoti hain wahan field strong hota hai; jahan woh phaili hoti hain wahan weak. Isiliye lines poles ke paas ghani hoti hain.
Yeh kabhi cross nahi karti. Crossing par $𝐁$ ko ek saath do directions mein point karna padta, jo impossible hai kyunki har point par field ki ek hi definite value hoti hai.

1/4Har line ek closed loop hai: N se bahar, bahar se ghoomkar, S mein andar aur magnet ke andar se wapas.

Fig. 5.4Bar magnet ki field lines: closed loops jinka tangent

𝐁

ki direction deta hai, jinki crowding uski strength batati hai, aur jo kabhi cross nahi karti.NCERT Fig. 5.2 · p. 138

Ek horizontal bar magnet, left mein S aur right mein N, magnet ke ek model se calculate ki gayi blue field lines ke saath. Lines N end se nikalti hain, upar aur neeche loop banati hain, S end mein ghusti hain aur magnet ke andar S se N tak chalti rehti hain. Step 1 ek poore loop ko orange mein highlight karta hai aur batata hai ki har line closed loop hai. Step 2 lines ko halka karta hai, unmein se ek ko highlight karta hai, aur us par ek point P mark karta hai jahan ek chhota compass hai jiski needle aur B label wala ek black tangent arrow line ke saath point karte hain. Step 3 do barabar square windows banata hai: N end ke paas wali window se chaar lines guzarti hain aur us par strong likha hai, top left mein door wali window se ek line guzarti hai aur us par weak likha hai. Step 4 ek inset dikhata hai jismein do lines cross karti hain, circle kiya hua aur impossible likha hua.

Upar wali activity ka compass method in lines ko banane ka sabse simple tareeka hai; iron filings poora pattern ek saath dikhati hain lekin yeh nahi dikha sakti ki lines kis taraf point karti hain. Simulation mein tum dono kar sakte ho.

SimulationBar magnet ka field map karna

Stage ke beech mein ek bar magnet jiski field lines bani hain, aur ek chhota compass jise tum kahin bhi drag kar sakte ho. Toggles se iron filings aur compass needles ka grid on hota hai. Readouts compass needle ki direction batate hain, aur yeh ki compass par field weak hai, moderate hai ya strong.

Fig. 5.5Compass ko lines ke saath drag karo, aur field lines, iron filings aur compasses ke grid ke beech switch karo.

Try karo

Iron filings on karo aur Field lines off karo. Kya filings wahi curves banati hain jo lines ne banaye the?
Compass ko ek field line ke saath N end se, bahar se ghoomte hue, S end tak dheere-dheere drag karo. Chalte hue needle kaise ghoomti hai?
Compass ko N end se bilkul sata kar rakho, phir use seedha bahar stage ke edge tak dheere-dheere drag karo. Field at compass reading kaise badalti hai, aur sabse tezi se kahan badalti hai?

Quick check

3 mein se 1

Bar magnet ke andar uski field lines kis taraf chalti hain?

S end se N end ki tarafN end se S end ki tarafMagnet ke andar woh hoti hi nahi

Answer dekho

Answer (A) S end se N end ki taraf

Bahar lines N se S jaati hain; loops close karne ke liye unhe magnet ke andar se S se N wapas aana padta hai. Woh poles par nahi rukti.

5.2.2 Equivalent solenoid ke roop mein bar magnet

Chapter 4 mein tumne dekha ki flat current loop ka magnetic moment $𝐦 = I 𝐀$ hota hai, aur $N$ turns wale solenoid ka $m = N I A$ uske axis ke saath hota hai. Door se aise solenoid ka field bilkul bar magnet ke field jaisa dikhta hai (Fig. 5.6). Ampère ek kadam aage gaye aur suggest kiya ki har magnet ka magnetism uske andar ke chhote circulating currents se aata hai. Aaj hum jaante hain ki woh currents kya hain: atoms mein electrons ka orbital motion aur spin.

3/3Electric dipole door se same pattern deta hai, lekin charges ke beech uski lines $+$ se $-$ chalti hain aur charges par khatam hoti hain.

Fig. 5.6Door se bar magnet, solenoid aur electric dipole lines ka same pattern banate hain. Andar

𝐁

𝐦

ke saath S se N chalta hai, lekin

𝐄

𝐩

ke against

+

-

chalta hai.NCERT Fig. 5.2 · p. 138

Ek hi picture ke teen versions jinki bahar wali field lines same hain. Step 1: ek short bar magnet, left mein S aur right mein N, blue field lines ke saath jo N se ghoomkar S tak loop banati hain aur magnet ke andar S se N chalti rehti hain, jise right ki taraf point karte orange arrowheads dikhate hain. Step 2: magnet ki jagah orange windings wala solenoid aata hai; field lines bilkul same hain, andar ke right wale arrows bhi. Step 3: source ek electric dipole ban jaata hai, left mein negative charge aur right mein positive charge, green electric field lines ke saath; door se unki shape pehle jaisi hi hai, lekin charges ke beech lines plus se minus chalti hain, isliye andar ka orange arrow left point karta hai.

Itne saare chhote-chhote loops milkar solenoid kaise ban sakte hain? Fig. 5.7 magnet ka end face chhote cells mein bata hua dikhata hai, har cell mein ek atomic current same direction mein ghoomta hai. Jahan do cells milte hain, wahan unke currents opposite directions mein chalte hain aur cancel ho jaate hain. Sirf bahar ke edge wale currents bachte hain, aur milkar surface ke chaaron taraf chalta ek bada loop banate hain. Magnet aisi slices ka stack hai, isliye woh apni hi surface par lapete solenoid jaisa behave karta hai. Ampère ki picture Fig. 5.1 wale cutting experiment ko bhi samjhati hai: do tukdon mein kati coil bas do chhoti coils hai.

3/4Sirf edge ke chaaron taraf wala current bachta hai: ek bada loop, coil ke ek turn jaisa.

Fig. 5.7Magnet ke andar paas-paas ke atomic currents cancel ho jaate hain, aur uski surface ke chaaron taraf ek current bachta hai. Asal mein magnet ek solenoid hai.

Step 1: magnet ka ek square end face four by four grid mein bata hai, har cell mein ek chhota blue current loop anticlockwise ghoom raha hai. Step 2: do paas ke cells ke beech ka common edge red circle se ghera gaya hai aur us par cancel likha hai, kyunki wahan dono currents opposite directions mein chalte hain. Step 3: andar ke loops halke ho jaate hain aur ek mota orange loop poore perimeter ke chaaron taraf anticlockwise chalta hai, yahi bacha hua surface current hai. Step 4: end view left khisakta hai aur right mein side view aata hai: ek bar jiska left S aur right N rang ka hai, orange windings mein lipta hai, front par current ke arrows neeche ki taraf hain, aur andar ek black arrow S se N point karta hai; end view par N face likha hai.

DerivationFinite solenoid ka far axial field

Length $2 l$ aur radius $a$ wala ek solenoid lo, jismein per unit length $n$ turns hain aur current $I$ behta hai. Uska centre origin par rakho aur uske axis par centre se $r$ doori par, door ke point P par field nikaalo: $r ≫ l$ aur $r ≫ a$ .

Position $x$ par $d x$ width ki patli slice mein $n d x$ turns hain, isliye yeh $n I d x$ current wali ek ring hai. Radius $a$ ki ring apne axis par doori $z$ par field $μ_{0} I a^{2} / 2 (a^{2} + z^{2})^{3 / 2}$ banati hai (Chapter 4). Yahan $z = r - x$ hai, aur kyunki $r - x ≫ a$ hai, hum $z^{2}$ ke saamne $a^{2}$ ko chhod sakte hain:

d B = \frac{μ_{0} n I a^{2} d x}{2 (r - x)^{3}}

Har slice axis ke saath same direction mein field deti hai, isliye hum unhe $x = - l$ se $x = + l$ tak jodte hain:

B = \frac{μ_{0} n I a^{2}}{2} \int_{- l}^{l} \frac{d x}{(r - x)^{3}} = \frac{μ_{0} n I a^{2}}{4} [\frac{1}{(r - l)^{2}} - \frac{1}{(r + l)^{2}}] = \frac{μ_{0} n I a^{2} r l}{(r^{2} - l^{2})^{2}}

Door $r ≫ l$ hai, isliye $(r^{2} - l^{2})^{2} \approx r^{4}$ aur $B \approx μ_{0} n I a^{2} l / r^{3}$ . Solenoid mein $n (2 l)$ turns hain, har ek ka area $π a^{2}$ , isliye uska magnetic moment $m = n (2 l) I π a^{2}$ hai, yaani $n I a^{2} l = m / 2 π$ . Isliye

Solenoid ya bar magnet ka far axial field

B = \frac{μ_{0}}{4 π} \frac{2 m}{r^{3}}

(5.1)

Isse do baatein nikalti hain. Pehli, real bar magnet ke axis ke saath chalaya gaya magnetometer, magnet se kuch lengths door jaate hi, yahi $1 / r^{3}$ fall-off paata hai, aur field N end se door point karta hai, isliye solenoid model experimental test pass karta hai. Doosri, isse humein ek working definition milti hai: bar magnet ko us solenoid ka moment $m$ diya jaata hai jo uska far field dobara bana de. Phir woh $m$ measure kiya ja sakta hai, jaise kisi known field mein needle ke swings ka time naapkar (Section 5.2.3).

Example

Bar magnet ban kar baitha solenoid

$6.0 c m$ lambe ek solenoid mein $1.2 c m$ radius ke 400 turns hain aur usmein $1.5 A$ current behta hai. (a) Uska magnetic moment nikaalo. (b) (5.1) se uske centre se $30 c m$ door axis par field nikaalo. (c) Finite solenoid ka exact calculation us point par $2.05 \times 10^{- 6} T$ deta hai. Approximation kitna achha hai, aur kyun?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

250 turns wali ek coil, har turn ka area

4.0 c m^{2}

, mein

0.60 A

current behta hai. Uska magnetic moment kya hai?

A m²

Answer dekho

Answer 0.06 A m²

m = N I A = 250 \times 0.60 \times 4.0 \times 10^{- 4} = 0.060 A m^{2}

c m^{2}

m^{2}

mein convert karna yaad rakho.

5.2.3 Uniform magnetic field mein dipole

Ek chhote magnet, jaise compass needle, ko uniform field $𝐁$ mein pivot karo (Fig. 5.8). Uska N end field ke saath dhakela jaata hai aur S end utni hi force se ulti taraf, isliye net force zero hai. Dono pushes ek hi line mein tabhi lagte hain jab needle field ke parallel ya antiparallel ho; warna woh ek couple banate hain, aur needle ek torque mehsoos karti hai jo use $𝐁$ ki taraf ghumata hai.

2/4 $θ = 60^{\circ}$ : forces ek couple banati hain, aur torque $m B \sin 60^{\circ}$ needle ko wapas field ki taraf ghumata hai.

Fig. 5.8Uniform field mein needle ke dono ends par forces cancel ho jaati hain, lekin jab tak needle field ke saath na ho, woh ek couple banati hain. Energy curve

U = - m B \cos θ

ki valley

0^{\circ}

par aur peak

180^{\circ}

par hai.NCERT Fig. 5.3(b) · p. 138

Left: right ki taraf point karti horizontal dashed field lines aur centre par pivoted ek needle, N end orange aur S end white. Green arrows dono ends par barabar aur opposite forces dikhate hain, N end wala field ke saath aur S end wala uske against. Right: angle theta ke against potential energy U ka graph, axis par minus 180 se plus 180 degrees mark hain: cosine shape ki ek valley jiska sabse neecha point zero par hai, aur ek dot current angle mark karta hai. Jab needle tirchi hoti hai, pivot se ek chhoti dashed line field direction mark karti hai aur ek arc theta mark karta hai. Step 1: needle field ke saath hai; forces ek line mein lagti hain, dot valley ke bottom par hai, us par stable likha hai. Step 2: needle 60 degrees tak ghumayi gayi hai; forces ek couple banati hain aur ek orange curved arrow dikhata hai ki torque use wapas field ki taraf ghuma raha hai; dot aadhe raaste upar hai. Step 3: needle 90 degrees par hai; torque sabse zyada hai aur dot U equals zero par hai. Step 4: needle field ke against point karti hai; forces phir ek line mein lagti hain, torque zero hai, aur dot curve ke top par hai, us par unstable likha hai.

DerivationMagnetic dipole ka torque aur potential energy

Current loop ke liye tumne Chapter 4 mein dikhaya tha ki uniform field mein torque $𝐦 \times 𝐁$ hota hai. Bar magnet ka moment uske equivalent solenoid jitna hi hota hai, isliye uske liye bhi yahi result lagta hai:

Magnetic dipole par torque

𝝉 = 𝐦 \times 𝐁, τ = m B \sin θ

(5.2)

jahan

$θ$: $𝐦$ aur $𝐁$ ke beech ka angle

Torque sabse zyada, $m B$ , tab hota hai jab needle field ke right angle par ho, aur zero tab jab woh field ke saath ya uske against ho. Yeh hamesha $θ$ ko kam karne ki koshish karta hai: yeh ek restoring torque hai.

Potential energy nikaalne ke liye needle ko angle $θ_{0}$ se $θ$ tak dheere-dheere ghumao, ek bahar ke torque se jo magnetic torque ko bas balance kare. Kiya gaya work potential energy ke roop mein store hota hai:

U (θ) - U (θ_{0}) = \int_{θ_{0}}^{θ} m B \sin θ^{'} d θ^{'} = - m B \cos θ + m B \cos θ_{0}

Sirf potential energy ke differences measure kiye ja sakte hain, isliye hum chun sakte hain ki $U = 0$ kahan ho. Jab needle field ke right angle par ho, $θ_{0} = 90^{\circ}$ , tab $U = 0$ chunne se constant hat jaata hai:

Magnetic dipole ki potential energy

U = - m B \cos θ = - 𝐦 \cdot 𝐁

(5.3)

$U$ sabse kam, $- m B$ , $θ = 0$ par hai. Field ke saath padi needle stable equilibrium mein hai: koi bhi chhota ghumaav uski energy badhata hai, aur torque use wapas le aata hai. $U$ sabse zyada, $+ m B$ , $θ = 180^{\circ}$ par hai. Field ke against point karti needle unstable equilibrium mein hai: wahan bhi torque zero hai, lekin zara sa dhakka badhta hi jaata hai.

DerivationSmall oscillations: B ya m measure karne ke liye needle ka time naapna (extension)

Needle ko field se thoda ghumao aur chhod do. Agar $I$ pivot ke around uska moment of inertia hai (yahan current nahi), to rotation ke liye Newton's second law deta hai

I \frac{d^{2} θ}{d t^{2}} = - m B \sin θ \approx - m B θ (θ chhota, radians mein)

Yeh angular frequency $ω = \sqrt{m B / I}$ wali simple harmonic motion ki equation hai, isliye needle is period se jhoolti hai

Small oscillations ka period

T = 2 π \sqrt{\frac{I}{m B}}

Known field mein swings ka time naapo, to tum $m$ nikaal sakte ho. Usi needle ka time unknown field mein naapo, to tum $B$ nikaal sakte ho, kyunki $B \propto 1 / T^{2}$ . Torque wala figure yahi arrangement dikhata hai: steady field mein jhoolne ke liye free needle.

InteractiveUniform field mein needle

Right ki taraf point karte uniform field mein pivoted ek needle, saath mein angle ke against potential energy ka graph jismein abhi ke angle par ek dot hai. Tum needle ko drag karke ya slider se angle set kar sakte ho, field ko 10 se 200 millitesla tak badal sakte ho aur magnetic moment chun sakte ho. Cards torque, potential energy, chhodne ke baad measured period aur small-swing formula dikhate hain. Buttons needle ko chhodte hain, rok kar rakhte hain, ya do degree ka nudge dete hain.

Fig. 5.9Angle, field aur magnetic moment set karo; torque aur energy padho; needle ko chhodo aur uske swings ka time naapo.

Example

Needle ka time naapna

$m = 5.0 \times 10^{- 2} A m^{2}$ aur moment of inertia $8.0 \times 10^{- 7} k g m^{2}$ wali ek magnetised needle horizontal plane mein jhool sakti hai. Jahan Earth ke field ka horizontal part $3.5 \times 10^{- 5} T$ hai: (a) uska small oscillations ka period nikaalo. (b) Ek electromagnet ke paas wahi needle $1.2 s$ ke period se jhoolti hai. Wahan total horizontal field ka estimate lagao, yeh maankar ki woh needle par uniform hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.1NCERT Example 5.1 · p. 139

Poles ki chaar paheliyan

(a) Loudspeaker ka flat ring magnet apni thickness ke through magnetised hota hai, isliye ek flat face N aur doosra S hota hai. Ise ek diameter ke across tod kar do C-shaped halves banaye jaate hain. Har half ke kaunse poles hote hain?

(b) Paani ke bowl mein ek cork par compass needle tair rahi hai. Woh ghoomkar north ki taraf ho jaati hai, lekin cork north ki taraf khiskta nahi. Jab bowl ke edge par ek chhota fridge magnet pakda jaata hai, to cork ghoomne ke saath-saath magnet ki taraf khiskta bhi hai. Is farq ko samjhao.

(c) Ek iron ring apni poori circumference ke saath magnetised hai, current wale toroid ke core ki tarah. Uske poles kahan hain?

(d) Bar magnet ka N pole steel rod ke dono ends ke paas baari-baari laaya jaata hai, aur dono ends attract hote hain. (i) Kya isse pata chalta hai ki rod magnetised hai? (ii) Ek doosri rod apne ek end par N pole ko repel karti hai. Isse kya pata chalta hai, aur is doosri rod ke kis hisse par ek chhoti latakti paper clip sabse zor se kheenchi jaayegi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

(b) Needle par Earth ka field practically uniform hai, isliye uske dono ends par pushes barabar aur opposite hain: torque hai lekin net force nahi. Fridge magnet ke paas field needle ki length ke saath bahut badalta hai. Needle aise ghoomti hai ki uska paas wala end magnet ke face ka opposite pole ban jaata hai, aur woh paas wala end door wale end se kahin zyada strong field mein hota hai. Paas wale end par attraction door wale end par repulsion se zyada hai, isliye magnet ki taraf ek net force lagta hai.
(d) (i) Nahi; asal mein isse lagta hai ki rod magnetised nahi hai. Magnet jo bhi end paas ho usmein opposite pole induce karta hai, isliye unmagnetised steel dono poles se attract hota hai. Magnetised rod ka ek end N pole ko repel karta, jab tak bar magnet itna strong na ho ki use dobara magnetise kar de. Attraction kuch prove nahi karta; sirf repulsion dikhata hai ki dono objects magnets hain. (ii) Repulsion ke liye like poles chahiye, isliye doosri rod ek magnet hai jiske us end par N pole hai. Clip uske do ends, yaani poles, ke paas sabse zor se kheenchi jaati hai, aur beech ke paas lagbhag bilkul nahi.

Quick check

4 mein se 1

𝐦

aur

𝐁

ke beech kis angle par magnet par torque sabse zyada hota hai?

0^{\circ}

45^{\circ}

90^{\circ}

180^{\circ}

Answer dekho

Answer (C) $90^{\circ}$

τ = m B \sin θ

tab sabse zyada hota hai jab

\sin θ = 1

0^{\circ}

aur

180^{\circ}

par torque zero hai, bhale hi

180^{\circ}

sabse zyada energy wali position hai.

5.2.4 Electrostatic analog

Ab tak ke results ko Chapter 1 ke electric dipole wale results ke saath rakho. Torque $𝐩 \times 𝐄$ ban jaata hai $𝐦 \times 𝐁$ , energy $- 𝐩 \cdot 𝐄$ ban jaati hai $- 𝐦 \cdot 𝐁$ , aur far axial field $2 p / 4 π ε_{0} r^{3}$ ban jaata hai $μ_{0} 2 m / 4 π r^{3}$ . Har magnetic formula apne electric partner se symbols ki ek simple adla-badli se nikal aata hai, isliye tum electric dipole ke saare jaane-pehchaane results dobara use kar sakte ho.

DerivationElectric-dipole results se short magnet ke fields

Replacements karo

𝐄 \to 𝐁, 𝐩 \to 𝐦, \frac{1}{4 π ε_{0}} \to \frac{μ_{0}}{4 π}

Electric dipole ke liye, uske size se kahin badi doori $r$ par, perpendicular bisector (equatorial line) par field $- 𝐩 / 4 π ε_{0} r^{3}$ hai, aur axis par $2 𝐩 / 4 π ε_{0} r^{3}$ . Swap karne se size $l ≪ r$ wale magnet ke fields milte hain:

Short magnet ka equatorial field

𝐁_{E} = - \frac{μ_{0}}{4 π} \frac{𝐦}{r^{3}}

(5.4)

Short magnet ka axial field

𝐁_{A} = \frac{μ_{0}}{4 π} \frac{2 𝐦}{r^{3}}

(5.5)

(5.5) ka size lo to tumhe (5.1) wapas mil jaata hai: solenoid calculation aur dipole analogy match karte hain, aur analogy direction bhi batati hai. Barabar doori par axial field equatorial field ka dugna hota hai; yeh $𝐦$ ke saath point karta hai, jabki equatorial field $𝐦$ ke opposite point karta hai (Fig. 5.10).

4/4Field lines directions confirm karti hain: N end ke aage $𝐦$ ke saath, magnet ke bagal mein $𝐦$ ke against.

Fig. 5.10Same doori

r

par short magnet ka axial field equatorial field ka dugna hota hai. Yeh

𝐦

ke saath point karta hai; equatorial field

𝐦

ke against point karta hai.NCERT p. 140

Ek short bar magnet, left mein S aur right mein N, halki dipole field lines ke saath. Axis par ek point magnet ke right mein r doori par hai aur equatorial line par doosra point utni hi doori r par upar hai, dono dashed lines se magnet se jude hain. Step 1 mein sirf magnet aur do points hain. Step 2 axial point par right ki taraf, magnet ke moment ke saath, ek lamba orange arrow jodta hai, jis par B sub A likha hai. Step 3 equatorial point par left ki taraf, moment ke opposite, ek chhota orange arrow jodta hai, jis par B sub E likha hai. Step 4 dono arrows aur ek box dikhata hai jismein ratio 2 to 1 likha hai.

Table 5.1Electric aur magnetic dipoles aamne-saamneNCERT Table 5.1 · p. 141

	Electric dipole	Magnetic dipole (short magnet)
Field law mein constant	$1 / 4 π ε_{0}$	$μ_{0} / 4 π$
Dipole moment aur uski unit	$𝐩$ , $C m$ mein	$𝐦$ , $A m^{2}$ mein
Axis par field ( $r ≫ l$ )	$2 𝐩 / 4 π ε_{0} r^{3}$	$μ_{0} 2 𝐦 / 4 π r^{3}$
Equatorial line par field	$- 𝐩 / 4 π ε_{0} r^{3}$	$- μ_{0} 𝐦 / 4 π r^{3}$
Uniform field mein torque	$𝐩 \times 𝐄$	$𝐦 \times 𝐁$
Uniform field mein energy	$- 𝐩 \cdot 𝐄$	$- 𝐦 \cdot 𝐁$
Uniform field mein net force	zero	zero
Source ke andar field lines	$+ q$ se $- q$ tak ( $𝐩$ ke against)	S se N tak ( $𝐦$ ke saath)
Ek end ke around surface se bahar flux	$q / ε_{0}$	zero

Example 5.2NCERT Example 5.2 · p. 141

Magnet ke paas compass needle

Moment $m_{A} = 0.20 A m^{2}$ wala ek chhota bar magnet A bench par fixed hai, uska N end east ki taraf hai. Moment $m_{C} = 0.050 A m^{2}$ wali compass needle C sirf apne pivot par ghoom sakti hai, aur compass ko chaar jagahon mein se kisi ek par rakha jaata hai, har jagah A ke centre se $r = 5.0 c m$ door hai: $C_{1}$ A ke axis par east mein, needle east ki taraf; $C_{2}$ axis par west mein, needle west ki taraf; $C_{3}$ perpendicular bisector par north mein, needle west ki taraf; $C_{4}$ bisector par south mein, needle north ki taraf. A aur C ko point dipoles maano aur Earth ka field ignore karo, jo yahan kai guna weak hai.

Fig. 5.11Magnet A (centre) aur compass C ki chaar jagahein, sab A se

r = 5.0 c m

door. Har needle ka orange end uska N end hai.

Ek dashed horizontal line magnet A ka axis aur ek dashed vertical line uska perpendicular bisector mark karti hai; kone mein chhote arrows north (upar) aur east (right) dikhate hain. Centre mein ek short bar magnet A hai, uska S half west mein aur N half east mein. Chaar gol compasses uske centre se barabar doori r par hain. Axis par, east wale compass C1 ki needle east ki taraf aur west wale C2 ki needle west ki taraf point karti hai. Bisector par, north wale C3 ki needle west ki taraf aur south wale C4 ki needle north ki taraf point karti hai.

(a) Har jagah par A ka field, magnitude aur direction, nikaalo. (b) Har jagah par needle ki potential energy nikaalo aur chaaron ko sabse kam se sabse zyada tak rank karo. (c) Kin jagahon par needle equilibrium mein hai, aur har equilibrium stable hai ya unstable? (d) Bachi hui jagah par needle par torque aur uske ghoomne ki direction nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Compass ko sahi rakhna

$m = 1.2 A m^{2}$ wala ek bar magnet ek compass ke saath same bench par rakha hai. Earth ke field ka horizontal part $4.0 \times 10^{- 5} T$ hai. Compass magnet se kitna door ho ki magnet ka field Earth ke field ka zyada se zyada 5 % rahe, agar compass (a) magnet ke axis par ho, (b) uski equatorial line par ho? Magnet ko short maano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Short magnet se same doori par axial field equatorial field ke comparison mein kaisa hai?

Size aur direction dono mein barabarAadha bada aur same taraf point karta haiDugna bada aur ulti taraf point karta hai

Answer dekho

Answer (C) Dugna bada aur ulti taraf point karta hai

B_{A} = μ_{0} 2 m / 4 π r^{3}

aur

B_{E} = μ_{0} m / 4 π r^{3}

, isliye

B_{A} = 2 B_{E}

𝐁_{A}

𝐦

ke saath point karta hai aur

𝐁_{E}

uske against.

5.3 Magnetism aur Gauss's law

Magnet ke N end ke chaaron taraf ek imaginary closed bag lapeto aur use cross karne wali field lines gino (Fig. 5.12). Electric dipole ke liye, positive charge ke chaaron taraf wahi bag mein ghusne se zyada lines nikalti hain, aur Chapter 1 ka Gauss's law us surplus ko andar ke charge, $q / ε_{0}$ , mein badal deta hai. Kya iske jaisa koi magnetic surplus hai?

2/4Lines surface ke bahar wale hisse se nikalti hain aur jahan surface magnet ko kaatti hai wahan ghusti hain: jitni nikalti hain utni hi ghusti hain.

Fig. 5.12Closed surface se nikalne wali har magnetic field line kahin na kahin usmein wapas ghusti hai, isliye kisi bhi closed surface se bahar magnetic flux zero hota hai. Andar ke charge se shuru hone wali electric lines bas bahar nikal jaati hain.NCERT Fig. 5.5 · p. 142

Left: ek bar magnet apni field lines ke saath, jinmein uske andar S se N chalti lines bhi hain. Step 1: ek dashed purple closed surface N end ko gherti hai aur magnet ko kaatti hai; us par ek chhota patch right mein bada karke dikhaya gaya hai, jismein patch, uska outward normal n aur normal se angle theta par ek field vector B hai. Step 2: har woh jagah mark hai jahan koi line surface cross karti hai, green dots jahan lines nikalti hain aur orange squares jahan ghusti hain; lines bahar wale hisse se nikalti hain aur magnet ke andar wale hisse se ghusti hain. Ek tally mein likha hai bahar 6, andar 6, net flux zero. Step 3: ek badi surface poore magnet ko gherti hai; phir se right mein 6 lines nikalti hain aur left mein 6 ghusti hain, net flux zero. Step 4: right mein ek positive charge dashed circle ke andar hai, aath radial electric field lines ke saath, sab bahar nikalti hui: bahar 8, andar 0.

Koi bhi closed surface banao, chahe woh magnet ke N end ko kaatti ho. Usse nikalne wali har field line ek closed loop ka hissa hai, isliye use kahin na kahin wapas ghusna hi padega. N end ke chaaron taraf lines surface ke bahar wale hisse se nikalti hain aur us hisse se ghusti hain jo magnet ko kaatta hai, jahan lines S se N chalti hain. Nikalne wali lines ki ginti hamesha ghusne wali lines ki ginti ke barabar hoti hai.

Ise exactly kehne ke liye closed surface $S$ ko chhote patches mein baanto aur har patch ko bahar ki taraf point karta area vector $Δ 𝐒$ do. Ek patch se magnetic flux $Δ ϕ_{B} = 𝐁 \cdot Δ 𝐒$ hai, jahan field nikalta hai wahan positive aur jahan ghusta hai wahan negative. Saare patches par jodne se zero aata hai:

Gauss's law for magnetism

ϕ_{B} = \sum_{saare patches} Δ ϕ_{B} = \sum_{saare patches} 𝐁 \cdot Δ 𝐒 = 0

(5.6)

LawGauss's law for magnetism

Tum koi bhi closed surface chuno, usmein jitna magnetic flux ghusta hai utna hi nikalta hai: $\oint_{S} 𝐁 \cdot d 𝐒 = 0$ .

Electric version, $\sum 𝐄 \cdot Δ 𝐒 = q / ε_{0}$ , ke right side mein charge hai kyunki charges hi woh jagah hain jahan electric lines shuru aur khatam hoti hain. Magnetic version ke right side mein zero hai kyunki nature $𝐁$ lines ke shuru ya khatam hone ke liye kuch nahi deti. Isliye magnetic field ka sabse chhota source koi ‘magnetic charge’ nahi balki ek current loop hai, jiska N face aur S face hamesha saath aate hain.

History seNCERT p. 142

Carl Friedrich Gauss (1777–1855)

Kehte hain ki school mein Gauss ne 1 se 100 tak ke numbers pal bhar mein jod diye the, aur 24 saal ki umar mein unhone theory of numbers par ek landmark book publish ki. 1801 mein unhone kuch hi observations se calculate kiya ki khoya hua dwarf planet Ceres kahan dobara dikhega, aur astronomers ne use wahin paaya. Göttingen mein unhone aur physicist Wilhelm Weber ne 1833 mein observatory aur physics laboratory ke beech ek wire chalayi aur uske through messages bheje, jo pehle electric telegraphs mein se ek tha. Dono ne kai deshon mein observers ko organise karke Earth ka magnetic field measure karwaya, aur Gauss ne dikhaya ki aise measurements ko length, mass aur time ki absolute units mein kaise express karte hain. Magnetic field ki CGS unit, gauss ( $1 G = 10^{- 4} T$ ), unke naam par hai.

Example 5.3NCERT Example 5.3 · p. 143

Impossible field lines pehchaano

Neeche ke chhe sketches kisi field ki lines dikhate hain. Har ek ke liye tay karo ki kya woh magnetic field lines dikha sakta hai. Agar nahi, to batao woh kaunsa rule todta hai, aur kya woh electrostatic field lines dikha sakta hai.

1/2Verdicts dekhne se pehle har sketch ke liye decide karo.

Fig. 5.13Inmein se kaunsi magnetic field lines ho sakti hain? Doosra step verdicts dikhata hai.

Chhe panels. A: ek chhoti ball se sab directions mein bahar nikalti seedhi lines. B: page se bahar current le jaati wire ke chaaron taraf concentric circles, arrows anticlockwise. C: ek bar magnet jiski bahar ki lines N se S chalti hain aur magnet ke andar ki lines bhi N se S banayi gayi hain. D: do lines jo ek doosre ko cross karti hain. E: ek solenoid jiske andar lagbhag seedhi parallel lines hain jo ends par phailti hain aur bahar mudkar wapas aati hain. F: khaali space mein ek akela closed oval loop jiske through kuch nahi guzarta. Doosre step mein B aur E ko green ticks aur A, C, D aur F ko orange crosses milte hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

C. Bahar ka hissa sahi hai, lekin magnet ke andar lines N se S ki taraf banayi gayi hain. Tab lines N end se bahar aur andar dono jagah nikalti hain, aur N end ke chaaron taraf surface se bahar flux zero nahi hai. Magnet ke andar $𝐁$ S se N chalta hai. $𝐁$ ke liye possible nahi. Lekin yeh ek sahi electrostatic picture hai: ek bar jiske right end par positive charge aur left end par negative charge ho, uska $𝐄$ bahar aur andar dono jagah $+$ se $-$ chalta hai, bilkul Fig. 5.6 ke electric dipole ki tarah.

Example 5.4NCERT Example 5.4 · p. 144

Field lines aur moments par chaar sawaal

(a) Electrons ki ek beam uniform magnetic field ki lines ke right angle par chhodi jaati hai. Kya woh lines ke saath chalti hai? Isse unhe lines of force kehne ke baare mein kya pata chalta hai?

(b) Maan lo strength $q_{m}$ ka ek magnetic monopole discover hota hai aur radius $R$ ke sphere ke centre mein trap kar liya jaata hai. Us sphere ke liye Gauss's law for magnetism kaisa dikhega, aur uski surface par tum kaisa field expect karoge?

(c) Orbit mein ghoomte spacecraft ke andar ek horseshoe magnet tair raha hai, kisi bhi doosre magnet se door. Har arm doosre arm ke banaye field mein hai. Kya magnet khiskna ya ghoomna shuru karega?

(d) Neutron par koi net charge nahi hota, phir bhi uska magnetic moment hota hai. Yeh kaise ho sakta hai, jab magnetic fields moving charges se aate hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek closed surface ek curved dome aur ek flat base se bani hai. Dome se nikalne wala magnetic flux

+ 3.6 \times 10^{- 5} W b

hai. Base se outward flux kitna hai, webers mein?

Answer dekho

Answer -0.000036 Wb

Total outward flux zero hai, isliye base par

- 3.6 \times 10^{- 5} W b

hona chahiye: utna hi flux usse andar ghusta hai.

5.4 Magnetisation aur magnetic intensity

Steel ki silaai wali sui aur steel ka girder barabar achhe se magnetise kiye ja sakte hain, phir bhi girder ka total magnetic moment laakhon guna bada hota hai, bas isliye ki woh bada hai. Kisi ek particular tukde ke bajaye material ko describe karne ke liye hum volume se divide karte hain. Moment khud atoms se aata hai: unke orbit karte aur spin karte electrons chhote current loops ki tarah kaam karte hain, aur sample ke andar atomic moments vectors ki tarah judte hain.

Definition

Magnetisation

Kisi material ka per unit volume net magnetic moment, $𝐌 = 𝐦_{net} / V$ . Yeh ek vector hai jo $A m^{2} / m^{3} = A / m$ mein measure hota hai, aur iske dimensions $[L^{- 1} A]$ hain.

Example $10 c m^{3}$ volume wali rod, jiske atomic moments milkar $2.0 A m^{2}$ dete hain, ka $M = 2.0 / (1.0 \times 10^{- 5}) = 2.0 \times 10^{5} A / m$ hota hai.

Magnetisation

𝐌 = \frac{𝐦_{net}}{V}

(5.7)

Magnetisation field ko kaise badalta hai, yeh dekhne ke liye ek lambe solenoid ko material se bhar do (Fig. 5.14).

3/4Moments line up ho jaate hain. Andar woh paas-paas ke pairs mein cancel ho jaate hain, aur core ki surface ke chaaron taraf ek magnetising current $I_{m}$ bachta hai.

Fig. 5.14Solenoid ko magnetised core se bharne par winding ke field

μ_{0} n I

mein ek field

μ_{0} 𝐌

jud jaata hai, maano core apni surface ke chaaron taraf extra current le ja raha ho.NCERT p. 145

Ek lambi coil tirchi orange windings ki tarah bani hai, aage wale turns par current ke arrows neeche ki taraf hain. Step 1: coil khaali hai aur teen blue field lines usmein se right ki taraf chalti hain. Step 2: ek pink core coil ko bharta hai; atomic magnetic moments ke chhote red arrows random directions mein point karte hain, isliye field nahi badalta. Step 3: saare atomic arrows ghoomkar field ke saath point karne lagte hain, aur core ki surface par neeche ki taraf arrows wali dashed purple lakeerein dikhti hain, yahi magnetising current hai jo extra windings ki tarah kaam karta hai. Step 4: andar ke field mein ab kahin zyada lines hain, aur atomic arrows background mein halke ho jaate hain.

DerivationBhare solenoid ke andar B, H aur M

Per unit length $n$ turns wale khaali lambe solenoid mein current $I$ behta ho, to uske andar

B_{0} = μ_{0} n I

(5.8)

Ab ise ek material se bhar do. Andar ka field winding ka field plus magnetised material ka ek contribution $𝐁_{m}$ ban jaata hai:

𝐁 = 𝐁_{0} + 𝐁_{m}

(5.9)

$𝐁_{m}$ , $𝐌$ ke proportional kyun hona chahiye? Jab atomic loops line up hote hain, to material ke andar paas-paas ke loops ke currents cancel ho jaate hain, bilkul Fig. 5.7 ki tarah, aur sirf surface ke chaaron taraf chalta current bachta hai. Thickness $d x$ aur cross-section $A$ wali core ki slice ka magnetic moment $M A d x$ hai, jo bilkul utna hi hai jitna uske chaaron taraf chalta surface current $M d x$ deta. Isliye $M$ ek current per unit length hai, jo winding mein $n I$ jaisa hi role play karta hai, aur jis reasoning se (5.8) mila tha usi se:

𝐁_{m} = μ_{0} 𝐌

(5.10)

Ab ek aisi quantity introduce karo jo material ke baare mein kuch jaane bina, sirf free currents se nikaali ja sake. Yeh magnetic intensity $𝐇$ hai, jise aise define karte hain

Magnetic intensity

𝐇 = \frac{𝐁}{μ_{0}} - 𝐌

(5.11)

taaki total field ho

𝐁 = μ_{0} (𝐇 + 𝐌)

(5.12)

Bhare solenoid mein $B = μ_{0} n I + μ_{0} M$ , isliye (5.11) se $H = n I$ milta hai. Magnetic intensity sirf free current par depend karti hai, core par nahi. Iski unit $M$ wali hi hai, ampere per metre.

Diya gaya $𝐇$ kitna magnetisation banata hai? Kai materials ke liye $𝐌$ seedha $𝐇$ ke proportional hota hai:

Magnetic susceptibility

𝐌 = χ 𝐇

(5.13)

Proportionality constant, ek pure number $χ$ , ko magnetic susceptibility (चुंबकीय प्रवृत्ति) kehte hain. Iska sign aur size materials ko families mein baant deta hai, jaisa Section 5.5 dikhata hai; negative $χ$ ka matlab hai ki $𝐌$ , $𝐇$ ke against point karta hai. (5.13) ko (5.12) mein daalne se milta hai

𝐁 = μ_{0} (1 + χ) 𝐇

(5.14)

Permeability

𝐁 = μ_{0} μ_{r} 𝐇 = μ 𝐇

(5.15)

jahan

$μ_{r} = 1 + χ$: relative permeability (ek pure number)
$μ = μ_{0} μ_{r}$: material ki permeability, $μ_{0}$ wali unit mein hi

Relative permeability $μ_{r}$ electrostatics ke dielectric constant $K$ ka magnetic cousin hai. Kyunki $μ_{r} = 1 + χ$ aur $μ = μ_{0} μ_{r}$ , data table mein har material ke liye $χ$ , $μ_{r}$ ya $μ$ mein se sirf ek dena kaafi hai.

Definition

Magnetic intensity

Vector $𝐇 = 𝐁 / μ_{0} - 𝐌$ , jo $A / m$ mein measure hota hai. Yeh sirf free currents se tay hota hai: lambe solenoid ke andar $H = n I$ , core chahe jo bhi ho.

Definition

Magnetic susceptibility

$H$ ke proportion mein magnetise hone wale material ke liye ratio $χ = M / H$ : diamagnets ke liye negative, paramagnets ke liye thoda positive, ferromagnets ke liye bahut bada.

InteractiveSolenoid core lab

Ek lambi coil jiska core tum air, bismuth, copper, water, aluminium, platinum, soft iron, ferrite aur superconducting lead mein se chun sakte ho. Tum turns per metre aur current set karte ho. Core se guzarti field lines dikhati hain ki B khaali coil ke comparison mein kaisa hai, aur chaho to chhote effects ko badha-chadha kar dikhaya ja sakta hai. Cards H, M, B, khaali coil ka field, susceptibility, relative permeability aur magnetising current dete hain, aur ek graph H ke against M plot karta hai jismein abhi ki value par ek dot hai.

Fig. 5.15Core, winding aur current chuno;

H

M

aur

B

dekho, aur

M

ke against

H

ka graph bhi.

Try karke dekho

Core lab mein try karne ki cheezein

1000 turns per metre aur $0.10 A$ current ke saath soft iron chuno. Khaali coil ke comparison mein $B$ lagbhag kitne guna bada hai?
Soft iron rakho aur current $5 A$ tak badhao. $μ_{r}$ ka kya hota hai, aur $M$ – $H$ graph kya dikhata hai?
Bismuth chuno. $H$ ke comparison mein $M$ kis taraf point karta hai, aur core $B$ ko kitne percent badalta hai?
Superconducting lead chuno. Uske andar $B$ aur $M$ kya hain?

$H = 100 A / m$ ke saath iron $B \approx 0.61 T$ deta hai, jo khaali coil ke $1.3 \times 10^{- 4} T$ ka lagbhag 5000 guna hai. $5 A$ par iron saturated ho jaata hai: $M$ lagbhag $1.7 \times 10^{6} A / m$ par ruk jaata hai, isliye $μ_{r}$ girkar kuch sau reh jaata hai. Real iron sirf chhote $H$ ke liye linear hota hai. Bismuth mein $M$ , $H$ ke opposite hai aur yeh $B$ ko sirf lagbhag $0.017 %$ kam karta hai. Superconductor mein andar $M = - H$ aur $B = 0$ hai.

Example 5.5NCERT Example 5.5 · p. 146

Ek ferrite ring

$8.0 c m$ mean radius wali ferrite ki ek ring par 400 turns barabar lapete gaye hain (Rowland ring). Winding mein $0.50 A$ current hone par ek search coil dikhati hai ki ferrite mein field $0.60 T$ hai. Nikaalo (a) ferrite mein $H$ , (b) $μ_{r}$ aur $χ$ , (c) $M$ , aur (d) magnetising current $I_{m}$ : woh extra current jo usi winding ko, bina ferrite ke, same $B$ dene ke liye chahiye hota.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Diamagnetism kitna chhota hai?

Ek bismuth rod ( $χ = - 1.66 \times 10^{- 4}$ ) 2000 turns per metre wale lambe solenoid ko bharti hai jismein $5.0 A$ current behta hai. Rod mein $H$ aur $M$ nikaalo, aur bismuth ki wajah se $B$ mein fractional change nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek lambe solenoid mein 1500 turns per metre hain aur

0.40 A

current behta hai. Uske andar

H

kitna hai, A/m mein?

A/m

Answer dekho

Answer 600 A/m

H = n I = 1500 \times 0.40 = 600 A / m

, solenoid mein chahe koi bhi material bhara ho.

5.5 Materials ki magnetic properties

$χ$ ke sign aur size ke peeche teen bilkul alag behaviours chhupe hain, aur yahi Table 5.2 ki teen families dete hain: diamagnetic, paramagnetic aur ferromagnetic. Table 5.3 ki values dikhati hain ki pehle do kitne weak hain, aksar size mein lagbhag $10^{- 5}$ . Phir bhi us chhote se number ka sign tay karta hai ki sample strong field se bahar dhakela jaayega ya uske andar kheencha jaayega.

Table 5.2Teen tarah ke magnetic materialsNCERT Table 5.2 · p. 147

	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility $χ$	$- 1 \leq χ < 0$ , aam taur par $- 10^{- 5}$ ke aas-paas	$0 < χ ≪ 1$ , aam taur par $10^{- 5}$ se $10^{- 3}$	$χ ≫ 1$ , saikdon se kai hazaar tak
Relative permeability $μ_{r}$	$0 \leq μ_{r} < 1$	1 se thoda zyada	$μ_{r} ≫ 1$
Permeability $μ$	$μ < μ_{0}$	$μ$ , $μ_{0}$ se thoda zyada	$μ ≫ μ_{0}$
Sample ke andar field	thoda weak	thoda strong	bahut zyada strong
Non-uniform field mein	weaker field ki taraf jaata hai	stronger field ki taraf halka sa jaata hai	stronger field ki taraf zor se kheencha jaata hai
Temperature ka effect	lagbhag koi nahi	temperature badhne par $χ$ girta hai	ek critical temperature ke upar paramagnetic ban jaata hai
Examples	bismuth, copper, gold, silver, water, nitrogen, zyadatar plastics	aluminium, platinum, magnesium, tungsten, oxygen gas, copper sulfate	iron, cobalt, nickel, gadolinium, alnico

Table 5.3Room temperature ke paas typical susceptibilities (SI units, lagbhag)

Material	Type	$χ$
bismuth	diamagnetic	$- 1.7 \times 10^{- 4}$
water	diamagnetic	$- 9.0 \times 10^{- 6}$
copper	diamagnetic	$- 9.7 \times 10^{- 6}$
air	paramagnetic	$+ 3.7 \times 10^{- 7}$
aluminium	paramagnetic	$+ 2.2 \times 10^{- 5}$
platinum	paramagnetic	$+ 2.7 \times 10^{- 4}$
soft iron	ferromagnetic	kuch hazaar ( $H$ par depend karta hai)

Fig. 5.16 yahi idea field lines se dikhata hai. Diamagnetic sample lines ko thoda side mein khiska deta hai, paramagnetic sample apne andar se kuch extra lines guzaar leta hai, aur ferromagnet unhe zor se apne andar kheench leta hai. Non-uniform field mein yeh farq forces ban jaate hain: paramagnets aur ferromagnets stronger field wale region ki taraf kheenche jaate hain, jabki diamagnets weaker field ki taraf dhakele jaate hain.

3/4Ferromagnet lines ko apne andar jama kar leta hai: andar field kai guna bada hota hai.

Fig. 5.16Sample uniform field ko kaise badalta hai: diamagnet use side mein dhakelta hai, paramagnet thoda andar kheenchta hai, ferromagnet zor se andar kheench leta hai aur superconductor use poori tarah bahar nikaal deta hai. Pehle do effects bahut badha-chadha kar dikhaye gaye hain.NCERT Fig. 5.7 · p. 147

Ek round sample uniform field mein rakha hai jiski lines left se right chalti hain; lines ek lambe cylinder ke liye calculate ki gayi hain. Step 1, diamagnetic: lines sample se thoda door mudti hain aur usmein se kam lines guzarti hain; effect badha-chadha kar dikhaya hai. Step 2, paramagnetic: lines sample ki taraf thoda mudti hain aur usmein se thodi zyada guzarti hain; yeh bhi badha-chadha kar. Step 3, ferromagnetic: lines zor se sample mein kheench li jaati hain, isliye picture ki lagbhag saari lines usmein se guzarti hain. Step 4, superconductor: koi line sample mein nahi ghusti; sab uske chaaron taraf mud jaati hain.

5.5.1 Diamagnetism

Strong magnet ke paas bismuth ka ek chhota tukda latkao, to woh thoda door khisakta hai, chahe tum koi bhi pole use karo. Aisa behave karne wale materials diamagnetic (प्रतिचुंबकीय पदार्थ) kehlaate hain. Unke andar field kam ho jaata hai, lekin bas thoda sa: water ya copper ke liye lagbhag $0.001 %$ .

Yeh repulsion aata kahan se hai? Diamagnetic atom mein electrons aise pairs mein hote hain ki unke magnetic moments cancel ho jaate hain; atom ka koi permanent moment nahi hota. Field on karne se har electron ke orbit se magnetic flux badalta hai, aur Lenz's law (Chapter 6) ke hisaab se induced effect us change ko oppose karta hai. Induced electric field dono orbits mein ek hi taraf ghoomta hai, isliye woh ek electron ko peeche kheenchta hai aur doosre ko aage dhakelta hai (Fig. 5.17). Dono moments ab cancel nahi hote: atom ko field ke opposite ek chhota moment mil jaata hai, aur isliye woh repel hota hai.

2/3Lenz's law se jis electron ka moment $𝐁$ ke saath point karta hai woh dheema hota hai, aur doosra tez.

Fig. 5.17Diamagnetic atom mein do opposite orbital moments cancel hote hain. Lagaya gaya field ek electron ko dheema aur doosre ko tez kar deta hai, aur

𝐁

ke opposite ek chhota moment bach jaata hai.

Thoda upar se dikhte do elliptical electron orbits, har ek ke centre mein nucleus aur orbit ke aage ek electron. Left orbit mein electron aage ki taraf left chalta hai aur uska magnetic moment arrow upar point karta hai; right orbit mein electron right chalta hai aur uska moment arrow neeche point karta hai. Step 1: koi field nahi; dono moment arrows barabar hain, isliye net moment zero hai. Step 2: ek upward field on hota hai, jise sides par dashed arrows dikhate hain; left electron dheema hota hai, uska velocity arrow aur upward moment chhote hote hain, jabki right electron tez hota hai aur uska downward moment badhta hai. Step 3: orbits ke beech ek purple arrow dikhata hai ki net moment neeche, field ke opposite, point karta hai.

Kyunki yeh effect orbit karte electrons se aata hai, jo har atom mein hote hain, koi bhi substance isse bach nahi sakta. Jahan para- ya ferromagnetism hota hai wahan yeh bas dab jaata hai, aur akela sirf un materials mein dikhta hai jinke atoms ka koi permanent moment nahi hota: jaise bismuth, copper, silver, gold, lead, water, nitrogen gas, common salt aur zyadatar organic compounds.

Lead ki ek disc ko lagbhag $7.2 K$ se neeche thanda karo aur ek chhote magnet ko uski taraf neeche lao: magnet ruk jaata hai aur hawa mein tairne lagta hai. Lead ab superconductor ban gaya hai. Uska electrical resistance gayab ho gaya hai, aur ab woh weak magnetic field ko apne andar bilkul nahi aane deta, isliye andar $B = 0$ , $χ = - 1$ aur $μ_{r} = 0$ . Is perfect diamagnetism ko Meissner effect kehte hain. Zero resistance aur zero internal field saath-saath kyun aate hain, is par physicists dashakon tak uljhe rahe. Jawab 1957 mein quantum mechanics se aaya: Bardeen, Cooper aur Schrieffer ki theory (BCS theory, Nobel Prize 1972) mein critical temperature ke neeche electrons pairs bana lete hain aur ek coordinated group ki tarah chalte hain. Superconducting coils MRI scanners aur particle accelerators chalate hain, aur magnetically levitated trains ko uthate hain.

Try karke dekho

Magnet se angoor ko dhakelo

Tumhe chahiye

lagbhag same size ke do angoor
ek plastic drinking straw
pivot ke liye ek bottle cap ya pencil ki tip
ek strong neodymium magnet
hawa ke jhonkon se bachane ke liye ek cardboard box

Pehle guess karo

Jab tum magnet ko ek angoor ke paas laoge to kya hoga?

Angoor dheere-dheere door dhakela jaata haiAngoor magnet ki taraf kheencha jaata haiKuch bhi nahi hota

Straw ke dono ends par ek-ek angoor laga do aur straw ko pivot par aise balance karo ki woh free hokar ghoome. Hawa rokne ke liye uske chaaron taraf box rakh do.
Straw ke ruk jaane tak intezaar karo. Magnet ke ek pole ko dheere-dheere ek angoor se kuch millimetre ki doori tak lao, use chhue bina.
Magnet ko wahan 20 se 30 seconds tak pakad kar rakho aur dekho. Phir doosre pole ke saath dohrao.

Quick check

3 mein se 1

Bismuth ka ek chhota tukda strong magnet ke ek pole ke paas latka hai. Kya hota hai?

Woh pole se door dhakela jaata hai, chahe koi bhi pole hoWoh pole ki taraf kheencha jaata haiWoh N se door dhakela jaata hai lekin S ki taraf kheencha jaata hai

Answer dekho

Answer (A) Woh pole se door dhakela jaata hai, chahe koi bhi pole ho

Bismuth diamagnetic hai: uska induced moment hamesha lagaye gaye field ko oppose karta hai, isliye woh weaker field ki taraf jaata hai, chahe koi bhi pole use ho. Ek pole se attraction aur doosre se repulsion permanent magnet ka behaviour hai.

5.5.2 Paramagnetism

Aluminium, platinum, oxygen aur iron, copper ya manganese ke kai compounds, jaise copper sulfate, strong magnet se halke se attract hote hain. In paramagnetic substances (अनुचुंबकीय पदार्थ) mein har atom, ion ya molecule khud ek chhota magnet hota hai, aam taur par isliye ki uske kuch electrons unpaired hote hain.

Jab koi field nahi hota, to atomic moments thermal motion ke dhakkon se har taraf point karte hain, aur average mein cancel ho jaate hain (Fig. 5.18). Lagaya gaya field unhe line up karne ki koshish karta hai jabki heat unhe bikherti rehti hai, isliye room temperature par sirf ek chhota extra hissa field ke saath point karta hai. Net effect $𝐇$ ke saath ek chhota magnetisation hai. Jaise aluminium ( $χ = 2.2 \times 10^{- 5}$ ) apne andar field ko sirf $0.002 %$ badhata hai. Itna weak hone par bhi yeh sample ko non-uniform field ke stronger hisse ki taraf kheenchne ke liye kaafi hai.

2/3Thermal agitation ke against field bas thoda sa jeetta hai: $𝐁_{0}$ ke saath ek weak magnetisation.

Fig. 5.18Thermal motion paramagnet ke permanent moments ko bikher deta hai. Field unhe thoda line up karta hai; stronger field ya kam temperature unhe aur line up karta hai, jab tak magnetisation saturate na ho jaaye.NCERT Fig. 5.7 · p. 148

Ek green rectangle sample ko dikhata hai aur usmein 24 chhote red arrows ka grid hai, har atomic moment ke liye ek, aur right mein M label wala ek vertical meter hai. Step 1: koi field nahi; arrows random directions mein point karte hain aur meter zero dikhata hai. Step 2: sample ke neeche right ki taraf point karta field arrow aata hai; arrows ab bhi bikhre hain lekin thoda right ki taraf jhukte hain, aur meter chhota magnetisation dikhata hai. Step 3: strong field aur kam temperature; lagbhag saare arrows right point karte hain aur meter lagbhag full hai.

Bahut zyada strong na hone wale fields ke liye susceptibility absolute temperature ke inversely proportional hoti hai, $χ = C / T$ , jahan $C$ material ka ek constant hai; yeh Curie's law hai. Yeh hamesha nahi chal sakta. Kuch kelvin par kai tesla ke fields mein lagbhag har atomic moment pehle se field ke saath point karta hai, isliye $M$ ke paas badhne ko kuch nahi bachta: woh apni saturation value par ruk jaata hai, jo ek atom ka moment guna per unit volume atoms ki sankhya hai.

Quick check

3 mein se 1

Jab koi external field nahi hota to paramagnetic sample mein magnetisation kyun nahi dikhta?

Uske atoms ke koi magnetic moments hote hi nahiUske atomic moments opposite pairs mein lock hote hainHeat uske atomic moments ko random point karte rakhti hai

Answer dekho

Answer (C) Heat uske atomic moments ko random point karte rakhti hai

Atoms ke permanent moments hote hain, lekin kisi ek direction ko favour karne wala field na ho to thermal agitation unhe random rakhta hai aur unka sum zero hota hai. Bina moments wale atoms, ya cancel karte pairs mein lock moments wale atoms, material ko diamagnetic banate.

5.5.3 Ferromagnetism

Jab iron aur aluminium dono ke atoms ke permanent moments hote hain, to iron itna alag kyun hai? Iron, cobalt, nickel aur gadolinium mein har atomic moment apne neighbours ko ek purely quantum-mechanical coupling se mehsoos karta hai jise exchange kehte hain, jiska koi classical explanation nahi hai. Yeh coupling neighbours ke same direction mein point karne ko favour karta hai, isliye moments bina kisi lagaye field ke, apne aap, domains naam ke regions mein line up ho jaate hain. Isiliye yeh metals magnet se zor se kheenche jaate hain aur khud permanent magnets ban sakte hain.

Ek domain aam taur par kuch micrometre se millimetre ke ek hisse tak bada hota hai; ek micrometre side wale cube mein bhi lagbhag $10^{11}$ iron atoms hote hain. Har domain ke andar magnetisation poora hota hai, lekin unmagnetised iron ke tukde mein paas ke domains alag-alag directions mein point karte hain aur sample ka koi net moment nahi hota (Fig. 5.19). Tum domains sach mein dekh sakte ho. Polish ki hui iron surface par bahut baareek magnetic powder wale liquid ki ek boond phailao, to powder domains ke beech ki walls ke saath jama ho jaata hai; microscope mein magnet paas laane par walls ko khiskte hue dekha ja sakta hai.

2/6Field ke saath wale domains apni walls khiskne se badhte hain.

Fig. 5.19Ferromagnet mein domains: lagaye gaye field mein badhte aur ghoomte hain, soft iron mein wapas bikhar jaate hain, hard magnet mein aligned rehte hain, aur Curie temperature ke upar gayab ho jaate hain.NCERT Fig. 5.8 · p. 149

Ferromagnet ka ek rectangular tukda irregular domains mein bata hai, har domain apne arrow ki direction ke hisaab se rangeen hai, aur right mein M label wala magnetisation meter hai. Step 1, koi field nahi: baarah domains barabar maatra mein upar, neeche, left aur right point karte hain aur meter zero dikhata hai. Step 2, right ki taraf weak field: right point karne wale domains apne neighbours ki keemat par bade ho gaye hain aur meter lagbhag aadha dikhata hai. Step 3, strong field: poora sample ek domain hai jismein har arrow right ki taraf hai aur meter full hai. Step 4, soft iron se field hataya: domains wapas mixed pattern mein aate hain aur meter lagbhag zero tak gir jaata hai. Step 5, hard magnet se field hataya: sample zyadatar aligned rehta hai, kuch chhote ulte domains ke saath, aur meter upar rehta hai. Step 6, Curie point ke upar garam kiya: domains gayab ho jaate hain aur chhote atomic arrows random directions mein point karte hain; meter zero dikhata hai.

Weak field lagao to domains ke beech ki walls khiskne lagti hain: jo domains pehle se lagbhag field ke saath point karte hain woh apne neighbours ko nigal jaate hain. Stronger field baaki domains ko ghuma deta hai, jab tak sample lagbhag field ke saath magnetised single domain na ban jaaye aur $M$ saturation ke paas na pahunch jaaye. Kyunki chhota $H$ itna bada $M$ banata hai, $μ_{r}$ hazaaron tak pahunchta hai (kuch special alloys ke liye $10^{5}$ se bhi upar). Yahi bahut bada magnetisation wajah hai ki nail ya paper clip magnet ke pole ki taraf jhapat kar kheench li jaati hai, jahan field sabse strong hota hai.

Field off karo, to dono tarah ke ferromagnets ke raaste alag ho jaate hain. Annealed iron jaise soft ferromagnet mein domain walls aasaani se khiskti hain, isliye domains wapas bikhre pattern mein aa jaate hain aur lagbhag koi magnetisation nahi bachta. Yeh electromagnets aur transformer cores ke liye sahi hai, jinhe badalte current ke saath chalna padta hai. Hard ferromagnet mein walls chhote grains aur impurities se atki hoti hain, isliye aligned pattern lock ho jaata hai. Alnico (aluminium, nickel aur cobalt ke saath iron ka alloy), neodymium–iron–boron aur mineral lodestone hard hain; inhi se speakers, motors, fridge doors aur compass needles ke permanent magnets bante hain.

Compare

Hard aur soft ferromagnets

Hard

field hatane par apna zyadatar magnetisation rakhta hai
magnetise karna mushkil aur demagnetise karna bhi mushkil
alnico, lodestone, neodymium–iron–boron
permanent magnets: compass needles, speakers, motors

Soft

field hatane par lagbhag saara magnetisation kho deta hai
magnetise aur demagnetise karna aasaan
soft iron, silicon steel, ferrites
electromagnets, relays, transformer cores

Heat order ko tod deti hai. Ferromagnet garam hone par thermal agitation uske domains ke andar alignment ko bigaadta hai aur magnetisation girta hai, pehle dheere aur phir tezi se. Ek critical temperature, Curie temperature (क्यूरी ताप), ke upar domains gayab ho jaate hain aur material paramagnetic ban jaata hai. Iron ke liye yeh lagbhag $1043 K$ (770 °C) par hota hai aur nickel ke liye lagbhag $627 K$ (354 °C) par. Gadolinium ke liye yeh sirf lagbhag $293 K$ (20 °C) hai, isliye gadolinium sirf thande din par hi ferromagnetic hota hai.

Quick check

4 mein se 1

Unmagnetised iron ke tukde ko magnetise karne ke in stages ko order mein lagao.

Domains kai directions mein point karte hain aur net magnetisation zero hai
Sample lagbhag ek domain hai aur magnetisation saturate ho jaata hai
Stronger field baaki domains ko field ki taraf ghumata hai
Weak field apne saath wale domains ko unki walls khiskakar bada karta hai

Answer dekho

Answer

Domains kai directions mein point karte hain aur net magnetisation zero hai
Weak field apne saath wale domains ko unki walls khiskakar bada karta hai
Stronger field baaki domains ko field ki taraf ghumata hai
Sample lagbhag ek domain hai aur magnetisation saturate ho jaata hai

Wall motion pehle aata hai kyunki usmein kam energy lagti hai; poore domains ghumane ke liye stronger field chahiye; saturation sabse aakhir mein aata hai.

Table 5.4Magnetic quantities ek nazar meinNCERT p. 151

Quantity	Symbol	Scalar ya vector	SI unit	Dimensions	Kaam ka relation
permeability of free space	$μ_{0}$	scalar	$T m / A$	$[M L T^{- 2} A^{- 2}]$	$μ_{0} / 4 π = 10^{- 7} T m / A$
magnetic field	$𝐁$	vector	$T$ (tesla)	$[M T^{- 2} A^{- 1}]$	$1 T = 10^{4} G$
magnetic moment	$𝐦$	vector	$A m^{2}$	$[L^{2} A]$	coil ke liye $m = N I A$
magnetic flux	$ϕ_{B}$	scalar	$W b$ (weber)	$[M L^{2} T^{- 2} A^{- 1}]$	$1 W b = 1 T m^{2}$
magnetisation	$𝐌$	vector	$A / m$	$[L^{- 1} A]$	per unit volume moment
magnetic intensity	$𝐇$	vector	$A / m$	$[L^{- 1} A]$	$𝐁 = μ_{0} (𝐇 + 𝐌)$
magnetic susceptibility	$χ$	scalar	koi nahi	koi nahi	$𝐌 = χ 𝐇$
relative permeability	$μ_{r}$	scalar	koi nahi	koi nahi	$μ_{r} = 1 + χ$
permeability	$μ$	scalar	$T m / A$	$[M L T^{- 2} A^{- 2}]$	$𝐁 = μ 𝐇$ , $μ = μ_{0} μ_{r}$

Recap

Summary aur key points

Magnetic poles pairs mein aate hain: magnet kaatne par chhote poore magnets milte hain, aur aaj tak koi magnetic monopole nahi mila (Section 5.1).
Magnetic field lines closed loops hoti hain (magnet ke andar S se N chalti hain); unka tangent $𝐁$ ki direction deta hai, unki crowding uski strength batati hai, aur woh kabhi cross nahi karti (Section 5.2.1).
Bar magnet solenoid ki tarah behave karta hai: atomic current loops milkar uski surface ke chaaron taraf ek current banate hain. Axis par door dono $B = μ_{0} 2 m / 4 π r^{3}$ dete hain (Section 5.2.2).
Uniform field mein dipole par koi net force nahi lagta lekin torque $𝝉 = 𝐦 \times 𝐁$ lagta hai. Uski energy $U = - 𝐦 \cdot 𝐁$ field ke saath sabse kam (stable) aur uske against sabse zyada (unstable) hoti hai (Section 5.2.3).
Moment of inertia $I$ wali needle period $T = 2 π \sqrt{I / m B}$ se jhoolti hai, jisse $m$ ya $B$ measure kiya ja sakta hai (Small oscillations ka period).
$𝐄 \to 𝐁$ , $𝐩 \to 𝐦$ aur $1 / 4 π ε_{0} \to μ_{0} / 4 π$ swap karne se electric-dipole results magnetic results ban jaate hain: $𝐁_{A} = μ_{0} 2 𝐦 / 4 π r^{3}$ aur $𝐁_{E} = - μ_{0} 𝐦 / 4 π r^{3}$ (Section 5.2.4).
Gauss's law for magnetism: kisi bhi closed surface par $𝐁$ ka flux milkar zero hota hai, kyunki sources ka kaam karne wale koi magnetic monopoles nahi hain (Section 5.3).
Magnetisation $𝐌$ per unit volume magnetic moment hai. Magnetic intensity $𝐇 = 𝐁 / μ_{0} - 𝐌$ free currents se tay hoti hai (solenoid mein $H = n I$ ), isliye $𝐁 = μ_{0} (𝐇 + 𝐌)$ (Section 5.4).
Linear material ke liye $𝐌 = χ 𝐇$ aur $𝐁 = μ_{0} μ_{r} 𝐇 = μ 𝐇$ , jahan $μ_{r} = 1 + χ$ ; $χ$ , $μ_{r}$ , $μ$ mein se koi ek baaki dono ko tay kar deta hai ((5.15)).
Diamagnets ( $χ$ chhota aur negative) weaker fields ki taraf dhakele jaate hain kyunki lagaya gaya field aise atomic moments induce karta hai jo use oppose karte hain; superconductor $χ = - 1$ wala perfect diamagnet hai (Section 5.5.1).
Paramagnets ( $χ$ chhota aur positive) mein permanent atomic moments hote hain jinhe field thermal agitation ke against thoda align karta hai; temperature girne par $χ$ badhta hai (Section 5.5.2).
Ferromagnets ( $χ ≫ 1$ ) mein domains hote hain jo field mein badhte aur ghoomte hain. Hard ferromagnets magnetised rehte hain aur soft nahi; Curie temperature ke upar sab paramagnetic ban jaate hain (Section 5.5.3).

Flash cards · 8

Yaad rakhne wale formulas

1 / 8

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 5.1	Practice Q1
NCERT Exercises 5.2	Practice Q2
NCERT Exercises 5.3	Practice Q3
NCERT Exercises 5.4	Practice Q4
NCERT Exercises 5.5	Practice Q5
NCERT Exercises 5.6	Practice Q6
NCERT Exercises 5.7	Practice Q7

Quick check: MCQs

Warm-up3 questions
5.2.1 Magnetic field lines3 questions
5.2.2 Equivalent solenoid ke roop mein bar magnet3 questions
5.2.3 Uniform magnetic field mein dipole4 questions
5.2.4 Electrostatic analog3 questions
5.3 Magnetism aur Gauss's law3 questions
5.4 Magnetisation aur magnetic intensity4 questions
5.5.1 Diamagnetism3 questions
5.5.2 Paramagnetism3 questions
5.5.3 Ferromagnetism4 questions

Glossary · 15

Key terms (English + Hindi)

Curie temperatureक्यूरी तापkyuri taap: Woh temperature jiske upar ferromagnet apne domains kho deta hai aur paramagnetic ban jaata hai: iron ke liye lagbhag $1043 K$ .Chapter mein pehli baar yahan aaya →

diamagneticप्रतिचुंबकीयpratichumbakiya: Chhota negative $χ$ : field ke against halka magnetise hota hai aur weaker fields ki taraf dhakela jaata hai.Chapter mein pehli baar yahan aaya →
domainडोमेनdomen: Ferromagnet ka ek region jismein saare atomic moments ek hi taraf point karte hain.Chapter mein pehli baar yahan aaya →

ferromagneticलौहचुंबकीयlauhchumbakiya: Bahut bada $χ$ , kyunki atomic moments domains mein line up hote hain; zor se attract hota hai aur permanent magnet ban sakta hai.Chapter mein pehli baar yahan aaya →

magnetic dipoleचुंबकीय द्विध्रुवchumbakiya dvidhruv: Barabar aur opposite poles ka pair, ya chhota current loop, jiska far field $1 / r^{3}$ ki tarah girta hai; bar magnet ek dipole hai.Chapter mein pehli baar yahan aaya →
magnetic field lineचुंबकीय क्षेत्र रेखाchumbakiya kshetra rekha: Ek curve jiska har point par tangent $𝐁$ ki direction deta hai; lines closed loops banati hain aur kabhi cross nahi karti.Chapter mein pehli baar yahan aaya →
magnetic intensityचुंबकीय तीव्रताchumbakiya teevrata: $𝐇 = 𝐁 / μ_{0} - 𝐌$ , sirf free currents se tay; lambe solenoid mein $H = n I$ .Chapter mein pehli baar yahan aaya →
magnetic momentचुंबकीय आघूर्णchumbakiya aaghurn: Vector $𝐦$ jo magnetic dipole ki strength aur direction naapta hai; coil ke liye $m = N I A$ , $A m^{2}$ mein.Chapter mein pehli baar yahan aaya →
magnetic monopoleचुंबकीय एकल ध्रुवchumbakiya ekal dhruv: Ek hypothetical akela N ya S pole. Aaj tak koi nahi mila, isiliye $\oint 𝐁 \cdot d 𝐒 = 0$ .Chapter mein pehli baar yahan aaya →
magnetic poleचुंबकीय ध्रुवchumbakiya dhruv: Magnet ka koi bhi end, jahan uski field lines nikalti hain (N pole) ya ghusti hain (S pole) aur jahan uska pull sabse strong hota hai.Chapter mein pehli baar yahan aaya →
magnetic susceptibilityचुंबकीय प्रवृत्तिchumbakiya pravritti: Dimensionless ratio $χ = M / H$ : diye gaye $H$ se material kitna strongly magnetise hota hai.Chapter mein pehli baar yahan aaya →
magnetisationचुंबकनchumbakan: Material ka per unit volume net magnetic moment, $𝐌$ , $A / m$ mein.
Meissner effectमाइसनर प्रभावmaisnar prabhav: Superconductor se weak magnetic field ka poori tarah bahar nikal jaana; superconductor $χ = - 1$ wale perfect diamagnet ki tarah behave karta hai.Chapter mein pehli baar yahan aaya →

paramagneticअनुचुंबकीयanuchumbakiya: Permanent atomic moments ki wajah se chhota positive $χ$ : stronger fields ki taraf halka kheencha jaata hai.Chapter mein pehli baar yahan aaya →

relative permeabilityआपेक्षिक चुंबकशीलताaapekshik chumbakasheelata: $μ_{r} = 1 + χ = μ / μ_{0}$ , woh factor jisse linear material $μ_{0} H$ ko multiply karta hai.Chapter mein pehli baar yahan aaya →

Magnetism and Matter: Class 12 Physics Chapter 5 Hinglish mein

Is chapter mein aap kya seekhenge

5.1 Introduction

5.2 Bar magnet

5.2.1 Magnetic field lines

5.2.2 Equivalent solenoid ke roop mein bar magnet

5.2.3 Uniform magnetic field mein dipole

5.2.4 Electrostatic analog

5.3 Magnetism aur Gauss's law

5.4 Magnetisation aur magnetic intensity

5.5 Materials ki magnetic properties

5.5.1 Diamagnetism

5.5.2 Paramagnetism

5.5.3 Ferromagnetism

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye